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4. オーストラリア規格 AS3600 スラブの設計例と SkyCiv との比較

# オーストラリア規格 AS3600 スラブの設計例と SkyCiv との比較

## 標準で考慮されるスラブシステム

オーストラリア規格は、鉄筋コンクリート スラブの設計に関する最小要件を定めています。, ワンウェイタイプ、ツーウェイタイプなど. 平面構成と梁の組み方について, スラブは、4 つの側面でサポートされているスラブに分割することもできます, 梁とスラブのシステム, フラットスラブ, と平板. これらのタイプは、次の画像にまとめられています.

• 句 6.10.2: 連続梁と一方向スラブ
• 句 6.10.3: 四方支持の二方スラブ
• 句 6.10.4: 複数のスパンを持つ双方向スラブ

コードの目的は、鉄筋の総量をスラブ システムの主な方向に設計することです。. Rebar steel will be calculated for the bending moments “Mx” そして “My.” 図 5 shows the other forces or actions in a finite slab element in which the code prescribes their resistance values.

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## One-Way Slab Design Example

Shown below is the small building and the slabs we will design

The plan dimensions are shown at next

For the slab example, in summary, the material, elements properties, and loads to consider :

• Slab type classification: 1 – way behaviour $$\フラク{L_2}{L_1} > 2 ; \フラク{14メートル}{6メートル}=2.33 > 2.00$$ OK!
• Building occupation: Residential use
• Slab thickness $$t_{スラブ}=0.25m$$
• Reinforced concrete density assuming a steel reinforcement ratio of 0.5% $$\rho_w = 24 \フラク{kN}{m^3} + 0.6 \フラク{kN}{m^3} \回 0.5 = 24.3 \フラク{kN}{m^3}$$
• Concrete characteristic compressive strength at 28 日々 $$f’c = 25 MPa$$
• Concrete Modulus of Elasticity by Australian Standard $$E_c = 26700 MPa$$
• Slab Self-Weight $$Dead = \rho_w \times t_{スラブ} = 24.3 \フラク{kN}{m^3} \times 0.25m = 6.075 \フラク {kN}{m^2}$$
• Super-imposed dead load $$SD = 3.0 \フラク {kN}{m^2}$$
• 活荷重 $$L = 2.0 \フラク {kN}{m^2}$$

### Hand calculation according to AS3600 Standard

このセクションで, we will calculate the required reinforced steel rebar using the reference of the Australian Standard. We first obtain the total factored bending moment to be carried out by the slab’s unitary width strip.

• Dead load, $$g = (3.0 + 6.075) \フラク{kN}{m^2} \回 1 m = 9.075 \フラク{kN}{メートル}$$
• 活荷重, $$q = (2.0) \フラク{kN}{m^2} \回 1 m = 2.0 \フラク{kN}{メートル}$$
• Ultimate load, $$Fd = 1.2\times g + 1.5\times q = (1.2\回 9.075 + 1.5\回 2.0)\フラク{kN}{メートル} =13.89 \frac{kN}{メートル}$$

Using the simplified method specified by the standard, 最初, it is a must to comply with the following restrictions:

• $$\フラク{L_i}{L_j} \インクルード 1.2 . \フラク{6メートル}{6メートル} =1 < 1.2$$. OK!
• Load has to be uniform. OK!
• $$q \le 2g. q=2 \frac{kN}{メートル} < 18.15 \フラク{kN}{メートル}$$. OK!
• The slab cross-section has to be uniform. OK!.

Recommended minimum thickness, d

$$d \ge \frac{L_{fe}}{{k_3}{k_4}{\平方根[3]{\フラク{\フラク{\デルタ}{L_{ef}}{E_c}}{F_{d, ef}}}}}$$

どこ

• $$k_3 = 1.0; k_4 = 1.75$$
• $$\フラク{\デルタ}{L_{ef}}=1/250$$
• $$E_c = 27600 MPa$$
• $$F_{d,ef} = (1.0 +k_{cs})\times g + (\psi_s + k_{cs}\times \psi_1) \times q=(1.0+0.8)\回 9.075 + (0.7+0.8\回 0.4)\回 2 = 18.375 kPa$$
• $$\psi_s = 0.7$$ Live-load short-term factor
• $$\psi_1 = 0.4$$ Live-load long-term factor
• $$k_{cs} = 0.8$$

$$d \ge \frac{5.50メートル}{{1.0}\回 {1.75}{\平方根[3]{\フラク{\フラク{1}{250}\回{27600 \times 10^3 kPa}}{18.375 kPa}}}} \ge 0.173m. d = 0.25m > 0.173メートル$$ OK!

Once we demonstrate that constraints are satisfied, the bending moment is calculated using the expression: $$M=\alpha \times F_d \times L_n^2$$ どこ $$\alpha$$ is a constant defined in the following figure.

どこ:

• (a) Case of slabs and beams on girder support
• (b) For continuous beam support only
• (c) Where Class L reinforcement is used
• $$L_n$$ is the unitary strip span
• $$F_d$$ is the gravitational factored load

For the slab example, we have to use case (a) because the slab rests on stiff girders. It will be explained only one case and the rest will show in the following table. We include also the steel reinforcement area calculation.

• $$M={\アルファ} {F_d}{L_n^2}={-\フラク{1}{24}}\回 {13.89 \フラク{kN}{メートル}}\回 (6メートル-0.5メートル)^2 = – 17.51{kN}{メートル}$$
• Cover = 20mm (A minimum of 10mm is needed for fire resistance period of 60 分).
• $$d = t_{スラブ} – カバー – \フラク{BarDiameter}{2} = 250mm – 20んん – 6mm = 224mm$$
• $$\alpha_2 = 1.0-0.003 f’c = 1.0-0.003\times 25 = 0.925 (0.67 \le \alpha_2 \le 0.85)$$ したがって, we select $$\alpha_2 = 0.85$$
• $$\xi = \frac{\alpha_2\times f’c}{f_{彼の}} = frac{0.85\回 25 MPa}{500 MPa} = 0.0425$$
• $$\rho_t = \xi – \平方根{{\xi}^ 2 – \フラク{{2}{\xi}{M}}{{\ファイ}{b}{d^2}{f_{彼の}}}} = 0.0425 – \平方根{{0.0425}^2-\frac{2\times 0.0425\times 17.51{kN}{メートル}}{{0.8}\回 {1メートル}\回 {{(0.224メートル)^ 2}} \回 {500\回 {10他のいくつかの例は}kPa}}}=0.0008814$$
• $$\gamma= 1.05-0.007 f’c = 1.05-0.007\times 25 = 0.875 (0.67 \le \gamma \le 0.85)$$ したがって, we select $$\gamma = 0.85$$
• $$k_u = \frac{\rho_t \times f_{彼の}}{0.85\times \gamma \times f’c}= frac{0.0008814\回 500 MPa}{0.85\回 0.85 \回 25 MPa} =0.0244$$
• $$\phi = 1.19 – \フラク{13\k_倍{u0}}{12} = 1.19 – \フラク{13\回 0.0244}{12} = 1.164 (0.6 \le \phi \le 0.8)$$ したがって, we select $$\phi = 0.8$$. OK!.
• $$\曲げモーメントは、セクションで各方向に計算されます{t,分} = 0.20 {(\フラク{D}{d})^ 2}{(\フラク{f’_{ct,f}}{f_{彼の}})} = 0.20 \回 (\フラク{0.25メートル}{0.224メートル})^2 \times \frac{0.6\回 sqrt{25MPa}}{500 MPa} = 0.0015$$
• $$A_{st}=max(\曲げモーメントは、セクションで各方向に計算されます{t,分}, \rho_t)\times b \times d = max(0.0015,0.0008814)\回 1000 mm \times 224 mm = 334.82 mm^2$$
$$\alpha$$ and Moments Exterior Negative Left Exterior Positive Exterior Negative Right Interior Negative Left Interior Positive Interior Negative Right
$$\alpha$$ 値 -$$\フラク{1}{24}$$ $$\フラク{1}{11}$$ -$$\フラク{1}{10}$$ $$\フラク{1}{10}$$ $$\フラク{1}{16}$$ $$\フラク{1}{11}$$
M value -17.51 38.20 -42.02 42.02 26.26 38.20
$$\rho_t$$ 0.0008814 0.001948 0.002148 0.002148 0.00133 0.001948
ku 0.0244 0.0539 0.0594 0.0594 0.0368 0.05391
$$\ファイ) 0.8 0.8 0.8 0.8 0.8 0.8 \(A_{st} {mm^2}$$ 334.82 436.31 481.099 481.099 334.8214 436.3100

After the steel rebar area calculation, you can define the detailing (the actual way to place the reinforcement into the slab). As help for your knowing, we share the following image, which indicates the rebar location for positive and negative moments:

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### SkyCiv S3D Plate Design Module Results

In the first view, we will show some images for the modeling and structural analysis of the example in S3D. We recommend you read about modeling in SkyCiv in the following links プレートのモデリング方法? そして ACI Slab Design Example with SkyCiv.

Before analyzing the model, we must define a plate mesh size. Some references (2) recommend a size for the shell element of 1/6 of the short span or 1/8 of the long span, the shorter of them. Following this value, 我々は持っています $$\フラク{L2}{6}= frac{6メートル}{6} = 1m$$ または $$\フラク{L1}{8}= frac{14メートル}{8}=1.75m$$; we take 1m as a maximum recommended size and 0.50m applied mesh size.

Once we improved our analytical structural model, we run a linear elastic analysis. When designing slabs, we have to check if the vertical displacement are less than the maximum allowed by code. Australian Standars stablished a maximum serviciability vertical displacement of $$\フラク{L}{250}= frac{6000んん}{250}=24.0 mm$$.

Comparing the maximimum vertical displacement against the code referenced value, the slab’s stiffness is adequate. $$4.822 んん < 24.00mm$$.

The maximum moments in the slab’s spans are located for positive in the center and for negative at the exterior and interior supports. Let’s see these moments values in the following images.

Plate element local axes are indicated below.

For more details about automated reinforced slab design, see our documentation Plates in SkyCiv.

### 結果比較

The last step in this one-way slab design example is compare the steel rebar area obtained by S3D analysis (local axes “2”) and handcalculations.

Moments and steel area Exterior Negative Left Exterior Positive Exterior Negative Right Interior Negative Left Interior Positive Interior Negative Right
$$A_{st, HandCalcs} {mm^2}$$ 334.82 436.31 481.099 481.099 334.8214 436.3100
$$A_{st, S3D} {mm^2}$$ 285.13 313.00 427.69 427.69 313.00 427.69
$$\デルタ_{dif}$$ (%) 14.84 28.262 11.101 11.101 6.517 1.986

We can see that the results of the values are very close to each other. This means the calculations are correct!

## Two-way Slab Design Example

このセクションで, we will develop an example that consists of a grillage system.

The plan dimensions are shown at next

For the slab example, in summary, the material, elements properties, and loads to consider :

• Slab type classification: Twoway behaviour $$\フラク{L_2}{L_1} \インクルード 2 ; \フラク{7メートル}{6メートル}=1.167 < 2.00$$ OK!
• Building occupation: Residential use
• Slab thickness $$t_{スラブ}=0.25m$$
• Reinforced concrete density assuming a steel reinforcement ratio of 0.5% $$\rho_w = 24 \フラク{kN}{m^3} + 0.6 \フラク{kN}{m^3} \回 0.5 = 24.3 \フラク{kN}{m^3}$$
• Concrete characteristic compressive strength at 28 日々 $$f’c = 25 MPa$$
• Concrete Modulus of Elasticity by Australian Standard $$E_c = 26700 MPa$$
• Slab Self-Weight $$Dead = \rho_w \times t_{スラブ} = 24.3 \フラク{kN}{m^3} \times 0.25m = 6.075 \フラク {kN}{m^2}$$
• Super-imposed dead load $$SD = 3.0 \フラク {kN}{m^2}$$
• 活荷重 $$L = 2.0 \フラク {kN}{m^2}$$

### Hand calculation according to AS3600 Standard

このセクションで, we will calculate the required reinforced steel rebar using the reference of the Australian Standard. We first obtain the total factored bending moment to be carried out by the slab’s unitary width strips in each bending main direction.

• Dead load, $$g = (3.0 + 6.075) \フラク{kN}{m^2} \回 1 m = 9.075 \フラク{kN}{メートル}$$
• 活荷重, $$q = (2.0) \フラク{kN}{m^2} \回 1 m = 2.0 \フラク{kN}{メートル}$$
• Ultimate load, $$Fd = 1.2\times g + 1.5\times q = (1.2\回 9.075 + 1.5\回 2.0)\フラク{kN}{メートル} =13.89 \frac{kN}{メートル}$$

Design moments and coefficients

Edge Condition Short-span coefficients ($$\beta_x$$) Long-span coefficients ($$\beta_y)$$ all values of $$\フラク{L_y}{L_x}$$
の値 $$\フラク{L_y}{L_x}$$
1.0 1.1 1.2 1.3 1.4 1.5 1.75 $$\の場合、ベースの下部から壁の高さの半分 2.0$$
1. Four edges continuous 0.024 0.028 0.032 0.035 0.037 0.040 0.044 0.048 0.024
2. One short edge discontinuos 0.028 0.032 0.036 0.038 0.041 0.043 0.047 0.050 0.028
3. One long edge discontinous 0.028 0.035 0.041 0.046 0.050 0.054 0.061 0.066 0.028
4. Two short edges discontinous 0.034 0.038 0.040 0.043 0.045 0.047 0.050 0.053 0.034
5. Two long edges discontinous 0.034 0.046 0.056 0.065 0.072 0.078 0.091 0.100 0.034
6. Two adjacent edges discontinous 0.035 0.041 0.046 0.051 0.055 0.058 0.065 0.070 0.035
7. Three edges discontinuous (one long edge continuous) 0.043 0.049 0.053 0.057 0.061 0.064 0.069 0.074 0.043
8. Three edges discontinuous (one short edge continous) 0.043 0.054 0.064 0.072 0.078 0.084 0.096 0.105 0.043
9. Four edges discontinuos 0.056 0.066 0.074 0.081 0.087 0.093 0.103 0.111 0.056

テーブル 1. (Yew-Chaye Loo & Sanual Hug Chowdhury , “補強およびプレストレストコンクリート”, 2nd edition, ケンブリッジ大学出版局)

The following image explain the all nine cases that the table above refers

Design moments for central region (場合 6 Two adjacent edges discontinuous) :

• $$L_x = 6m, L_y=7m, \フラク{L_y}{L_x} = frac{7メートル}{6メートル}= 1.167$$ Values to be linearly interpolated
• Positives:
• $$M_x = {\beta_x}{F_d}{L_x^2} = {0.04435}\回 {13.89 \フラク{kN}{メートル}}\回{(6メートル)^ 2}=22.177 kNm$$
• $$M_y = {\beta_y}{F_d}{L_x^2} ={0.035}\回 {13.89 \フラク{kN}{メートル}}\回{(6メートル)^ 2}=17.501 kNm$$
• Negatives exterior span:
• $$M_{x1,A} = -\lambda_e \times M_x = -0.5 \回 22.177 kNm = – 11.089 kNm$$
• $$M_{y1,A} = -\lambda_e \times M_y = -0.5 \回 17.501 kNm = -8.751 kNm$$
• Negatives interior span:
• $$M_{x1,B} = -\lambda_{1バツ} \times M_x = -1.33 \回 22.177 kNm = – 29.495 kNm$$
• $$M_{y1, B} = -\lambda_{1そして} \times M_y = -1.33 \回 17.501 kNm = -23.276 kNm$$

Design moments for central region (場合 3 One long edge discontinous) :

• $$L_x = 6m, L_y=7m, \フラク{L_y}{L_x} = frac{7メートル}{6メートル}= 1.167$$ Values to be linearly interpolated
• Positives:
• $$M_x = {\beta_x}{F_d}{L_x^2} = {0.03902}\回 {13.89 \フラク{kN}{メートル}}\回{(6メートル)^ 2}= 19.512 kNm$$
• $$M_y = {\beta_y}{F_d}{L_x^2} ={0.028}\回 {13.89 \フラク{kN}{メートル}}\回{(6メートル)^ 2}= 14.001 kNm$$
• Negatives interior span:
• $$M_{x1,B} = -\lambda_{1バツ} \times M_x = -1.33 \回 19.512 kNm = – 25.951 kNm$$
• $$M_{y1,B} = -\lambda_{1そして} \times M_y = -1.33 \回 14.001 kNm = – 18.621 kNm$$
• Negatives interior second span:
• $$M_{x2,B} = -\lambda_{2バツ} \times M_x = -1.33 \回 19.512 kNm = – 25.951 kNm$$
• $$M_{y2,B} = -\lambda_{2そして} \times M_y = -1.33 \回 14.001 kNm = – 18.621 kNm$$

Rebar steel for X direction

$$\alpha$$ and Moments Exterior Negative Left Exterior Positive Exterior Negative Right Interior Negative Left Interior Positive Interior Negative Right
M value 11.089 22.177 29.495 25.951 19.512 25.951
$$\rho_t$$ 0.00055614 0.00112 0.001496 0.001313 0.000984 0.001313
ku 0.015395 0.0310 0.0414 0.0364 0.0272 0.0364
$$\ファイ) 0.8 0.8 0.8 0.8 0.8 0.8 \(A_{st} {mm^2}$$ 334.8214 334.8214 335.08233 334.821 334.8214 334.8214

Rebar steel for Y direction

$$\alpha$$ and Moments Exterior Negative Left Exterior Positive Exterior Negative Right Interior Negative Left Interior Positive Interior Negative Right
M value 8.751 17.501 23.276 18.621 14.001 18.621
$$\rho_t$$ 0.0004383 0.0008811 0.001176 0.0009381 0.000703 0.0009381
ku 0.0121 0.0244 0.03256 0.02597 0.0195 0.02597
$$\ファイ) 0.8 0.8 0.8 0.8 0.8 0.8 \(A_{st} {mm^2}$$ 334.821 334.821 334.821 334.821 334.8214 334.821

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### SkyCiv S3D Plate Design Module Results

After refining the model, is time to run a linear elastic analysis.

When designing slabs, we have to check if the vertical displacement are less than the maximum allowed by code. Australian Standars stablished a maximum serviciability vertical displacement of $$\フラク{L}{250}= frac{6000んん}{250}=24.0 mm$$.

The image above gaves to us the vertical displacement. The maximum value is -1.179mm being less than the maximum allowed of -24mm. したがって, the slab’s stiffeness is adequate.

Images 27 そして 28 consist of the bending moment in each main direction. Taking the moment distribution and values, the software, SkyCiv, can obtain then the total steel reinforcement area.

Steel reinforcement areas:

### 結果比較

The last step in this one-way slab design example is compare the steel rebar area obtained by S3D analysis and handcalculations.

Rebar steel for X direction

Moments and steel area Exterior Negative Left Exterior Positive Exterior Negative Right Interior Negative Left Interior Positive Interior Negative Right
$$A_{st, HandCalcs} {mm^2}$$ 334.8214 334.8214 335.08233 334.821 334.8214 334.8214
$$A_{st, S3D} {mm^2}$$ 289.75 149.35 325.967 325.967 116.16 217.311
$$\デルタ_{dif}$$ (%) 13.461 55.39 2.720 2.644 65.307 35.0964

Rebar steel for Y direction

Moments and steel area Exterior Negative Left Exterior Positive Exterior Negative Right Interior Negative Left Interior Positive Interior Negative Right
$$A_{st, HandCalcs} {mm^2}$$ 334.821 334.821 334.821 334.821 334.821 334.821
$$A_{st, S3D} {mm^2}$$ 270.524 156.75 304.34 304.34 156.75 270.52
$$\デルタ_{dif}$$ (%) 19.203 53.184 9.104 9.104 53.184 19.204

The diference is some high for positive moments and the reason would be the presence of beams with high torsional stiffness that impact on the Plate Finite Element Analysis Results and the calculations for bending reinforcement steel.

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## 参考文献

1. Yew-Chaye Loo & Sanual Hug Chowdhury , “補強およびプレストレストコンクリート”, 2nd edition, ケンブリッジ大学出版局.
2. Bazan Enrique & Meli Piralla, “Diseño Sísmico de Estructuras”, 1ed, LIMUSA.
3. オーストラリア規格, コンクリート構造物, なので 3600:2018

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